Forcing $2^{\omega} = \omega_{\omega_1}$ together with $2^{\omega_1} = \omega_{\omega_2}$

Question

I'm studying from Kunen's Set Theory, and I came across this exercise (G6 ch. 7):

Suppose $M$ satisfies $\text{GCH}$. Let $\kappa_1 < \dots < \kappa_n$ be regular cardinals of $M$ and let $\lambda_1 \le \lambda_2 \le\dots\le\lambda_n$ be cardinals of $M$ such that $(\text{cof}(\lambda_i)>\kappa_i)^M$. Force $n$ times to construct a c.t.m (countable transitive model) $N\supset M$ with the same cardinals such that for each $i$, $(2^{\kappa_i} = \lambda_i)^N$.

Now, I have some problems in proving this in its generality. I'll exemplify this:
suppose I want to show the general statement above for $n = 2, \kappa_1 = \omega, \kappa_2 = \omega_1, \lambda_1 = \omega_{\omega_1}, \lambda_2 = \omega_{\omega_2}$.

I would first use the forcing notion $\mathbb{P}_1 = \text{Fn}(\omega_{\omega_2}\times \omega_1, 2, \omega_1)$. In fact, by taking a $\mathbb{P}_1$-general filter $G$ over M, we have that $M[G]$ preserves cardinals and, thanks to $\text{GCH}$ in $M$, we also have $(2^{\omega_1} = \omega_{\omega_2})^{M[G]}$.

Secondly I'd use (inside $M[G]$) the forcing notion $\mathbb{P}_2 = \text{Fn}(\omega_{\omega_1}\times \omega, 2, \omega)$. Since $(\mathbb{P}_2 \text{ ccc})^{M[G]}$, this second forcing preserves cardinals. A problem arises when we want to show that $(2^\omega = \omega_{\omega_1})^{M[G][Q]}$ with $Q$ a $\mathbb{P}_2$-generic filter over $M[G]$.
Of course we have $(2^\omega \ge \omega_{\omega_1})^{M[G][Q]}$ but to show also $(2^\omega \le \omega_{\omega_1})^{M[G][Q]}$ we need to deal with nice names in $\mathbb{P}_2$. By doing so we get $(2^{\omega} \le \theta)^{M[G][Q]}$ with $(\theta = \omega_{\omega_1}^\omega)^{M[G]}$. So we are left to prove $(\omega_{\omega_1}^\omega \le \omega_{\omega_1})^{M[G]}$. But I don't see how to do this.

If we were to continue , we would still need show that $(2^{\omega_1} = \omega_{\omega_2})^{M[G][Q]}$:

of course I have $(2^{\omega_1} \ge \omega_{\omega_2})^{M[G][Q]}$ since we are preserving cardinals and this holds trivially in $M[G]$.
To show that $(2^{\omega_1} \le \omega_{\omega_2})^{M[G][Q]}$ I'd again use the nice names within $\mathbb{P}_2$, and this gives us $(2^{\omega_1} \le \theta)^{M[G][Q]}$ where $(\theta = \omega_{\omega_1}^{\omega_1})^{M[G]}$.

Hence the final argument would be to prove that $(\omega_{\omega_1}^{\omega_1} \le \omega_{\omega_2})^{M[G]}$, but again I don't see how to do this. Any hint?

Thanks!

EDIT: I noticed that the problem was arising also in the midst of the proof, so I added it and simply rescaled everything by one cardinals to make it "simpler".

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EDIT2: If you don't want to understand my try, it's ok of course. I'd be obliged if you just mention how you would prove it, regardless of what I did! I'm referring to the original exercise I quoted or the more specific one I proposed.

EDIT3: As someone pointed out, the first forcing is $\omega_1$-closed, hence we have $(\omega_{\omega_1}^\omega)^M = (\omega_{\omega_1}^\omega)^{M[G]}$. Therefore the first thing is solved. It remains the second one (i.e. $(\omega_{\omega_1}^{\omega_1} \le \omega_{\omega_2})^{M[G]}$)

Thanks again

Answer 1

Since the first forcing is $\omega_1$-closed, there are no new mappings $\omega\to M,$ so the set of mappings $\omega\to \omega_{\omega_1}$ is the same in $M[G]$ as in $M$ and so since cardinals are preserved, $$(\omega_{\omega_1}^{\omega})^{M[G]} = (\omega_{\omega_1}^{\omega})^{M}=\omega_{\omega_1}.$$

For the other missing part, just observe that in $M[G]$, $$ \omega_{\omega_1} <\omega_{\omega_2} = 2^{\omega_1},$$ so $$ \omega_{\omega_1}^{\omega_1} = 2^{\omega_1} = \omega_{\omega_2}.$$